An insurance policy is written to cover a loss X where X has a density function
$$f(x) = \frac{3}{8}x^2; \qquad 0 \le x \le 2$$
The time (in hours) to process a claim of size x, where $0 \le x \le 2$, is uniformly distributed on the interval from x to 2x.
Calculate the probability that a randomly chosen claim on this policy is processed in three hours or more.
My Work
The problem is asking me to solve for $P[T>3]$. I'll first find the joint density function, and then integrate over the appropriate area.
$f(x,t) = f(t\mid X=x) \dot f(x) = \frac{3x^2}{8}\frac{1}{x}$ $= \frac{3x}{8}$
A graph of the area
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This leads me to $P[T>3] = \int_3^4 \int_{1.5}^2 \frac{3x}{8} \,dx\,dy = .328$
however the correct answer is 0.17. where have I gone wrong?
Best Answer
The former parts are all right, except you didn't pay attention to the definition region of $t$.
The joint distribution of $t$ and $x$ is: $$ f(x,t)=f_{0}(t\mid x)f(x)=3x/8~~~1\leq x\leq 2, ~x\leq t \leq 2x $$ From the graph you draw, we could see the integration should be: $$P[T>3]=\int_3^4 dt \int_{t/2}^{2}3x/8 \, dx=0.171875$$